Question: You have found the following ages (in years) of all 4 turtles at your local zoo: $ 22,\enspace 111,\enspace 26,\enspace 10$ What is the average age of the turtles at your zoo? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we have data for all 4 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population variance $({\sigma^2})$ To find the population mean , add up the values of all $4$ ages and divide by $4$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\mu} = \dfrac{22 + 111 + 26 + 10}{{4}} = {42.3\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $-20.3$ years $412.09$ years $^2$ $111$ years $68.7$ years $4719.69$ years $^2$ $26$ years $-16.3$ years $265.69$ years $^2$ $10$ years $-32.3$ years $1043.29$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{412.09} + {4719.69} + {265.69} + {1043.29}} {{4}} $ $ {\sigma^2} = \dfrac{{6440.76}}{{4}} = {1610.19\text{ years}^2} $ The average turtle at the zoo is 42.3 years old. The population variance is 1610.19 years $^2$.